1. ## Simple geormetry/congruence help

ABCD is a quadrilateral with AB horizontal and CD vertical. Angle BAD = angle BCD = 40 degrees and AB=BC=5. Show that angle CDB = angle ADB = 25 degrees.

Here is a diagram I made which I hope is correct:

http://img443.imageshack.us/img443/4404/geom1.jpg

By observing alternate angles, I can see that angle ADC = 50 degrees but how do I show that angle CDB = angle ADB? I thought about proving congruence but is this possible here?

2. ## Re: Simple geormetry/congruence help

Let's analize triangles ADB and BDC:
- angle DAB = angle BCD
- BD is a common lature
- AB = BC = 5
Rezults from the case of congruence side-side-angle triangle ADB= triangle BDC so angleADB=angleBDC

AB is horizontal and CD is vertical, so if we extend the side AB we'll have a right angle where AB intersect CD (point F). In triangle AFD, F=90 degrees, A=40 degrees, so angleADF = angleADC = 180-90-40 = 50 degrees.

3. ## Re: Simple geormetry/congruence help

Originally Posted by draganicimw
Let's analize triangles ADB and BDC:
- angle DAB = angle BCD
- BD is a common lature
- AB = BC = 5
Rezults from the case of congruence side-side-angle triangle ADB= triangle BDC so angleADB=angleBDC

AB is horizontal and CD is vertical, so if we extend the side AB we'll have a right angle where AB intersect CD (point F). In triangle AFD, F=90 degrees, A=40 degrees, so angleADF = angleADC = 180-90-40 = 50 degrees.